An industrial electronics technician is servicing a timing circuit on an automated assembl… — Armed Services Vocational Aptitude Battery (ASVAB)

An industrial electronics technician is servicing a timing circuit on an automated assembly line. The circuit network contains two $12\text{ }\mu\text{F}$ capacitors connected in series, and this combination is connected in parallel with a third capacitor rated at $14\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this circuit network in microfarads ( $\mu\text{F}$ )?

Answer: 20 μF

Answer

20\text{ }\mu\text{F}

For the series branch, two identical

12\text{ }\mu\text{F}

capacitors yield an equivalent capacitance of

6\text{ }\mu\text{F}

. When placed in parallel with the

14\text{ }\mu\text{F}

capacitor, the capacitances add directly (

6\text{ }\mu\text{F} + 14\text{ }\mu\text{F}

), giving a total equivalent capacitance of

20\text{ }\mu\text{F}

Step-by-Step Solution

Calculate the equivalent capacitance of the two

12\text{ }\mu\text{F}

capacitors connected in series.

C_s = 6\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}

, or

C_s = \frac{C}{n} = \frac{12}{2} = 6\text{ }\mu\text{F}

Add the capacitance of the parallel branch to the series equivalent.

C_{\text{total}} = 20\text{ }\mu\text{F}

Capacitors in parallel add directly together:

C_{\text{total}} = C_s + C_3 = 6\text{ }\mu\text{F} + 14\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

Key Concept

Series and parallel capacitor combinations and equivalent capacitance calculations

Estimated Time:1m 0s

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